How is capacitance related to the distance between the plates of a capacitor?

What is the wavelength of an ekectromagnetic wave that has a frequency of 40,000 hz? the speed of light is 3x10^8 m/s?

ehidna [41]

Solution
As,
speed=frequency × wavelengths
so,
wavelength=speed/frequency
=3×10^8/40000
=7500m

7 0

3 years ago

Water flows through a first pipe of diameter 3 inches. If it is desired to use another pipe for the same flow rate such that the

Alborosie

Answer:

the diameter of the second pipe is 2.52 in

Explanation:

Given the data in the question;

We know that; the rate of flow is the same;

so

Av1 = Av2

v ∝ √h

$\frac{A1}{A2}$ = $\frac{V2}{V1}$

$\frac{A1}{A2}$ = √( $\frac{h2}{h1}$ )

( π/4.D1² / π/4.D2² ) = √( $\frac{h2}{h1}$ )

( D1² / D2² ) = √( $\frac{2h1}{h1}$ ) since second is double of first

so

( D1² / D2² ) = √( $\frac{2}{1}$ )

3² / D2² = √2

D2²√2 = 9

D2² = 9/√2

D2² = 9 / 1.4142

D2² = 6.364

D2 = √ 6.364

D2 = 2.52 in

Therefore, the diameter of the second pipe is 2.52 in

3 0

3 years ago

Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m

Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0

3 years ago

Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket's engine is to be fired

ValentinkaMS [17]

Answer:

$\frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=1096.633$

Explanation:

$m_i$ = Initial mass of rocket

$m_f$ = Final mass of rocket

u = Initial velocity

$v_r$ = Relative velocity

v = Velocity

From the rocket equation

$v=u+v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v}}=e^1\\\Rightarrow \frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=2.7182$

when $v=7v_r$

$v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{7v_r}{v_r}}=e^7\\\Rightarrow \frac{m_i}{m_f}=1096.633$

$\frac{m_i}{m_f}=1096.633$

6 0

3 years ago

You jump off a platform 134 m above the Neuse River. After you have free-fallen for the first 40 m, the bungee cord attached to

omeli [17]

Answer:

The acceleration at lowest point is 19.62 m/s^2

Explanation:

Conservation of energy is an concept in which it is stated that the energy of an isolated object remains the same. Energy changes from one form to another.

Lets Assume

Constant of string is K

By using the conservation of energy we will have the following equation

1/2 x 80^2 x K = m x 9.81 x 120

3200 K = 1177.2 m

K = 1177.2 m / 3200

K = 0.368 m

At the lowest point we will have

a = ( K x X - m x g ) / m

a = ( 0.368 m x 80 - m x 9.81 ) / m

a = 19.62 m / s^2

So, the acceleration at lowest point is 19.62 m/s^2

7 0

3 years ago

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